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Hwksolns10TA

# Hwksolns10TA - Solutions to HW 10 Problems and Exercises...

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Solutions to HW 10 Problems and Exercises: 37.28. Visualize: At t t t 0 s, the origins of the S, S , and S reference frames coincide. Solve: We have 1 1 2 2 2 2 1 ( / ) 1 (0.80) 1.667. v c (a) Using the Lorentz transformations, 8 ( ) 1.667 1200 m (0.80)(3 10 m/s)(2.0 10 s) x x vt 6 1200 m 8 6 2 2 8 (0.80)(3 10 m/s)(1200 m) 1.667 2.0 10 s 2.0 s 3 10 m/s vx t t c     (b) Using v  c , the above equations yield  x 2800 m and 8.67 s. t   37.29. Solve: We have 1 1 2 2 2 2 1 1 0.60 1.25. v c In the earth’s reference frame, the We assume that the axes of the two frames coincide at t=t’= 0 sec Lorentz transformations yield  10 8 10 10 8 10 2 2 8 ( ) 1.25 3.0 10 m 0.60 3.0 10 m/s 200 8.25 10 m 8.3 10 m 0.60 3.0 10 m/s 3.0 10 m 1.25 200 325 s 330 s 3.0 10 m/s x x vt v x t t c 37.30. Model: S is the ground’s frame and S is the rocket’s frame. S moves with velocity v 0.5 c relative to S. Solve: (a) We have 1 1 2 2 2 2 1 / 1 0.50 1.155. v c Applying the Lorentz transformations to the lightning strike at x 0 m and t 10 s,  8 5 5 2 (1.155) 0 m 0.5 3.0 10 m/s 1 10 s 1732 m 1700 m (1.155)(1 10 s 0 s) 11.55 s 12 s x x vt vx t t c         For the lightning strike at x 30 km and t 10 s,   4 8 5 8 4 5 2 8 1.155 3.0 10 m 0.50 3.0 10 m/s 1 10 s 32.91 m 33 m 0.50 3.0 10 m/s 3.0 10 m 1.155 1 10 s 46.2 s 46 s 3.0 10 m/s x t         (b) The events in the rocket’s frame are not simultaneous. The lightning is observed to strike the pole before the tree by 46 12 58 s. Hence we could infer, from direct application of Lorentz transformation (which is less confusing and more universal ), that simultaneity is relative.

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